Wiki
Wiki
Courses and public working groups
Courses and public working groups
Private Working Groups
Private Working Groups
- New!!! Reading Group
- Theatre
- Admin
- Research
- Teaching
Sidebar
world:measure
$$ \newcommand{\arginf}{\mathrm{arginf}} \newcommand{\argmin}{\mathrm{argmin}} \newcommand{\argmax}{\mathrm{argmax}} \newcommand{\asconv}[1]{\stackrel{#1-a.s.}{\rightarrow}} \newcommand{\Aset}{\mathsf{A}} \newcommand{\b}[1]{{\mathbf{#1}}} \newcommand{\ball}[1]{\mathsf{B}(#1)} \newcommand{\bbQ}{{\mathbb Q}} \newcommand{\bproof}{\textbf{Proof :}\quad} \newcommand{\bmuf}[2]{b_{#1,#2}} \newcommand{\card}{\mathrm{card}} \newcommand{\chunk}[3]{{#1}_{#2:#3}} \newcommand{\condtrans}[3]{p_{#1}(#2|#3)} \newcommand{\convprob}[1]{\stackrel{#1-\text{prob}}{\rightarrow}} \newcommand{\Cov}{\mathbb{C}\mathrm{ov}} \newcommand{\cro}[1]{\langle #1 \rangle} \newcommand{\CPE}[2]{\PE\lr{#1| #2}} \renewcommand{\det}{\mathrm{det}} \newcommand{\dimlabel}{\mathsf{m}} \newcommand{\dimU}{\mathsf{q}} \newcommand{\dimX}{\mathsf{d}} \newcommand{\dimY}{\mathsf{p}} \newcommand{\dlim}{\Rightarrow} \newcommand{\e}[1]{{\left\lfloor #1 \right\rfloor}} \newcommand{\eproof}{\quad \Box} \newcommand{\eremark}{</WRAP>} \newcommand{\eqdef}{:=} \newcommand{\eqlaw}{\stackrel{\mathcal{L}}{=}} \newcommand{\eqsp}{\;} \newcommand{\Eset}{ {\mathsf E}} \newcommand{\esssup}{\mathrm{essup}} \newcommand{\fr}[1]{{\left\langle #1 \right\rangle}} \newcommand{\falph}{f} \renewcommand{\geq}{\geqslant} \newcommand{\hchi}{\hat \chi} \newcommand{\Hset}{\mathsf{H}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\img}{\text{Im}} \newcommand{\indi}[1]{\mathbf{1}_{#1}} \newcommand{\indiacc}[1]{\mathbf{1}_{\{#1\}}} \newcommand{\indin}[1]{\mathbf{1}\{#1\}} \newcommand{\itemm}{\quad \quad \blacktriangleright \;} \newcommand{\jointtrans}[3]{p_{#1}(#2,#3)} \newcommand{\ker}{\text{Ker}} \newcommand{\klbck}[2]{\mathrm{K}\lr{#1||#2}} \newcommand{\law}{\mathcal{L}} \newcommand{\labelinit}{\pi} \newcommand{\labelkernel}{Q} \renewcommand{\leq}{\leqslant} \newcommand{\lone}{\mathsf{L}_1} \newcommand{\lrav}[1]{\left|#1 \right|} \newcommand{\lr}[1]{\left(#1 \right)} \newcommand{\lrb}[1]{\left[#1 \right]} \newcommand{\lrc}[1]{\left\{#1 \right\}} \newcommand{\lrcb}[1]{\left\{#1 \right\}} \newcommand{\ltwo}[1]{\PE^{1/2}\lrb{\lrcb{#1}^2}} \newcommand{\Ltwo}{\mathrm{L}^2} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mcbb}{\mathcal B} \newcommand{\mcf}{\mathcal{F}} \newcommand{\meas}[1]{\mathrm{M}_{#1}} \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\normmat}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \newcommand{\nset}{\mathbb N} \newcommand{\N}{\mathcal{N}} \newcommand{\one}{\mathsf{1}} \newcommand{\PE}{\mathbb E} \newcommand{\pminfty}{_{-\infty}^\infty} \newcommand{\PP}{\mathbb P} \newcommand{\projorth}[1]{\mathsf{P}^\perp_{#1}} \newcommand{\Psif}{\Psi_f} \newcommand{\pscal}[2]{\langle #1,#2\rangle} \newcommand{\pscal}[2]{\langle #1,#2\rangle} \newcommand{\psconv}{\stackrel{\PP-a.s.}{\rightarrow}} \newcommand{\qset}{\mathbb Q} \newcommand{\revcondtrans}[3]{q_{#1}(#2|#3)} \newcommand{\rmd}{\mathrm d} \newcommand{\rme}{\mathrm e} \newcommand{\rmi}{\mathrm i} \newcommand{\Rset}{\mathbb{R}} \newcommand{\rset}{\mathbb{R}} \newcommand{\rti}{\sigma} \newcommand{\section}[1]{==== #1 ====} \newcommand{\seq}[2]{\lrc{#1\eqsp: \eqsp #2}} \newcommand{\set}[2]{\lrc{#1\eqsp: \eqsp #2}} \newcommand{\sg}{\mathrm{sgn}} \newcommand{\supnorm}[1]{\left\|#1\right\|_{\infty}} \newcommand{\thv}{{\theta_\star}} \newcommand{\tmu}{ {\tilde{\mu}}} \newcommand{\Tset}{ {\mathsf{T}}} \newcommand{\Tsigma}{ {\mathcal{T}}} \newcommand{\ttheta}{{\tilde \theta}} \newcommand{\tv}[1]{\left\|#1\right\|_{\mathrm{TV}}} \newcommand{\unif}{\mathrm{Unif}} \newcommand{\weaklim}[1]{\stackrel{\mathcal{L}_{#1}}{\rightsquigarrow}} \newcommand{\Xset}{{\mathsf X}} \newcommand{\Xsigma}{\mathcal X} \newcommand{\Yset}{{\mathsf Y}} \newcommand{\Ysigma}{\mathcal Y} \newcommand{\Var}{\mathbb{V}\mathrm{ar}} \newcommand{\zset}{\mathbb{Z}} \newcommand{\Zset}{\mathsf{Z}} $$
This result is taken from Billiingsley, Probability and measure , 3rd edition, page 46.
Statement
There exists no probability measure $P$ on $\mathcal{P}(\rset)$such that $P\{x\} = 0$ for each $x \in \mathbb{R}$.
Proof
The proof of this impossibility theorem requires the well-ordering principle (equivalent to the axiom of choice) and also the continuum hypothesis. Let $S$ be the set of sequences $(s(1), s(2), \ldots)$ of positive integers. Then $S$ has the power of the continuum. (Let the $n$th partial sum of a sequence in $S$ be the position of the $n$th $1$ in the nonterminating dyadic representation of a point in $(0, 1]$; this gives a one-to-one correspondence.) By the continuum hypothesis, the elements of $S$ can be put in a one-to-one correspondence with the set of ordinals preceding the first uncountable ordinal. Carrying the well-ordering of these ordinals over to $S$ by means of the correspondence gives $S$ a well-ordering relation $<_w$ with the property that each element has only countably many predecessors.
For $s, t \in S$, write $s < t$ if $s(i) < t(i)$ for all $i > 1$. Say that $t$ rejects $s$ if $t <_w s$ and $s < t$; this is a transitive relation. Let $T$ be the set of unrejected elements of $S$. Let $V_s$ be the set of elements that reject $s$, and assume it is nonempty. If $t$ is the first element (with respect to $<_w$) of $V_s$, then $t \in T$ (if $t'$ rejects $t$, then it also rejects $s$, and since $t <_w t'$, there is a contradiction). Therefore, if $s$ is rejected at all, it is rejected by an element of $T$.
Suppose $T$ is countable, and let $t_1, t_2, \ldots$ be an enumeration of its elements. If $t^*(k) = t_k(k) + 1$, then $t^*$ is not rejected by any $t_k$ and hence lies in $T$, which is impossible because it is distinct from each $t_k$. Thus, $T$ is uncountable and must, by the continuum hypothesis, have the power of $(0, 1]$.
Let $x$ be a one-to-one map of $T$ onto $(0, 1]$; write the image of $1$ as $x_1$. Let $A_k = \{x : t(i) = k\}$ be the image under $x$ of the set of $t$ in $T$ for which $t(i) = k$. Since $t(i)$ must have some value $k$, $\bigcup_{k=1}^{\infty} A_k = (0, 1]$. Assume that $P$ is countably additive and choose $u$ in $S$ in such a way that $P\left(\bigcup_{k=1}^{\infty} A_k\right) > 1 - \frac{1}{2^i+1}$ for $i > 1$. If \[ A = \bigcap_{k=1}^{\infty} A_k = \bigcap_{k=1}^{\infty} \{x : t(i) < u(i)\} = \{ x : t < u \}, \] then $P(A) > 0$. If $A$ is shown to be countable, this will contradict the hypothesis that each singleton has probability $0$.
Now, there is some $t_0$ in $T$ such that $u < t_0$ (if $u \in T$, take $t_0 = u$; otherwise, $u$ is rejected by some $t_0$ in $T$). If $t < u$ for $t \in T$, then $t < t_0$, and hence $t \not\leq_w t_0$ (since otherwise $t_0$ rejects $t$). This means that $\{t : t < u\}$ is contained in the countable set $\{t : t \leq_w t_0\}$, and $A$ is indeed countable.
world/measure.txt · Last modified: 2023/10/07 12:32 by rdouc
